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Fed [463]
2 years ago
11

HELP !!show the work tooo

Mathematics
1 answer:
Elden [556K]2 years ago
8 0

the probability of getting heads on a two-face coin is simply 1/2.

now, what is the probability of getting 1/2 AND 1/2 AND 1/2 AND 1/2 AND 1/2?   well, AND means "times", namely

\cfrac{1}{2}\stackrel{and}{\cdot }\cfrac{1}{2}\stackrel{and}{\cdot }\cfrac{1}{2}\stackrel{and}{\cdot }\cfrac{1}{2}\stackrel{and}{\cdot }\cfrac{1}{2}\implies \cfrac{1}{32}

well, and since it's a two-face coin, the probability for each face is equal, 1/2, so Heads has a chance of 1/2 each time, or we can also say that Tails has a chance of 1/2 each time, so if Tails has the same probability over 5 times as Head does.

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| 2(4+6) = how do I solve this?​
podryga [215]

Answer:

20

Step-by-step explanation:

You use the distributive property.

Distribute the 2 to whatever is inside the parentheses. (You will multiply)

4×2=8

2×6=12

Substitute the answers above into the parentheses.

(8+12)=20

7 0
3 years ago
Which expressions are equivalent to 2(b+3c)?
Rina8888 [55]

Answer:

B. (b+3c)+(b+3c)

C. 2(b)+2(3c)

Step-by-step explanation:

we have

2(b+3c)

Distribute the number 2

2(b+3c)=2b+2(3c)=2b+6c

Verify each case

case A) 3(b+2c)

distribute the number 3

3(b+2c)=3b+3(2c)=3b+6c

3b+6c \neq 2b+6c

therefore

Choice A is not equivalent to the given expression

case B) (b+3c)+(b+3c)

Combine like terms

b+3c)+(b+3c)=(b+b)=(3c+3c)=2b+6c

2b+6c= 2b+6c

therefore

Choice B is equivalent to the given expression

case C) 2(b)+2(3c)

Multiply both terns by 2

2(b)+2(3c)=2b+6c

2b+6c= 2b+6c

therefore

Choice C is equivalent to the given expression

6 0
3 years ago
Read 2 more answers
Which of these expressions is equivalent to 6x – 10x + 20?
erastova [34]

Answer:

4(5-x)

Step-by-step explanation:

6x-10x+20         4(5-x)

-4x+20                20-4x

and those two are both equivalent

4 0
3 years ago
HELPPP IM SOOO STUCKK PLEASE!
Taya2010 [7]

Answer:

60% that his or her drink will be a medium.

Step-by-step explanation:

Total amount orders = 5 + 48 + 22 +8 +12 +5

= 100

48 + 12 = 60 (medium orders)

60 out of 100 = 60%

Therefore it is 60%

8 0
4 years ago
In a certain school there are 180 pupils in the year7,110 pupils study french,88 study German , and 65 indonesian.40 pupils stud
MAXImum [283]

Answer:

(a) All three languages  - 9

(b) Indonesian only  - 10

(c) none of the languages - 12

(d) At least one language  - 168

(e) Either one or two of the three languages - 159​

Step-by-step explanation:

The question require the knowledge of Set theory and its Formulas.

Total pupils = 180

French (Let F) = 110  

German (Let G) = 88

Indonesian (Let I) =  65

French and German (F intersection G) =  40

German and Indonesian (G intersection I) =  38

French and Indonesian (F intersection I)  = 26

German only  = 19

(a) All three languages

We are given only German speaking people are 19

Only German = n(G) - n( F intersection G) - n(G intersection I) + n(F intersection G intersection I)

 19 = 88 - 40-38 + n(G intersection I) + n(F intersection G intersection I)

n(G intersection I) + n(F intersection G intersection I) = 9

n(G intersection I) + n(F intersection G intersection I) represents the number of pupils speaking who study all the three languages.

(b)Indonesian only

   n( I) - n(G intersection I) + n(F intersection I) + n(G intersection I) + n(F intersection G intersection I)

65 - 38 - 26 + 9 = 10

So 10 pupils speak Indonesian only

(c)none of the languages

It will be equal to Total - pupils speaking any of the three languages

Any of the three languages = (F union G union I)

  = n(F) +n(G) + n(I) -n(F intersection G) - n(G intersection I) -n(F intersection I)

    +  n(F intersection G intersection I)

n(G intersection I)

  = 110 + 88 + 65 -40 -38-26 + 9

  = 263 - 95

   = 168

So 168 pupils speak any of the three languages

None speakers = 180 - 168 = 12 pupils

So 12 pupils do not speak any of the three languages.

(d)at least one language

At least one language has the meaning that the person can either speak one two or all three languages, so it will be same as we proceeded above

Any of the three languages = (F union G union I)

  = n(F) +n(G) + n(I) -n(F intersection G) - n(G intersection I) -n(F intersection I)

    +  n(F intersection G intersection I)

n(G intersection I)

  = 110 + 88 + 65 -40 -38-26 + 9

  = 263 - 95

   = 168

(e) either one or two of the three languages​

Pupils can speak one or two language but not all the three so will subtract all the three language speaker from the total.

  Total speaker = 168

 All three languages speaker  = 9

 Either one or two of the three languages​ speaker = 168 - 9 = 159

8 0
4 years ago
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